Solve x^2+(14-x)^2=8^2

$$x ^ { 2 } + ( 14 - x ) ^ { 2 } = 8 ^ { 2 }$$

Solve for x (complex solution)

$$x=7+\sqrt{17}i\approx 7+4.123105626i$$

$$x=-\sqrt{17}i+7\approx 7-4.123105626i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(14-x\right)^{2}.

$$x^{2}+196-28x+x^{2}=8^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+196-28x=8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$2x^{2}+196-28x=64$$

Subtract 64 from both sides.

$$2x^{2}+196-28x-64=0$$

Subtract 64 from 196 to get 132.

$$2x^{2}+132-28x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-28x+132=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -28 for b, and 132 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 2\times 132}}{2\times 2}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 2\times 132}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-28\right)±\sqrt{784-8\times 132}}{2\times 2}$$

Multiply -8 times 132.

$$x=\frac{-\left(-28\right)±\sqrt{784-1056}}{2\times 2}$$

Add 784 to -1056.

$$x=\frac{-\left(-28\right)±\sqrt{-272}}{2\times 2}$$

Take the square root of -272.

$$x=\frac{-\left(-28\right)±4\sqrt{17}i}{2\times 2}$$

The opposite of -28 is 28.

$$x=\frac{28±4\sqrt{17}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{28±4\sqrt{17}i}{4}$$

Now solve the equation x=\frac{28±4\sqrt{17}i}{4} when ± is plus. Add 28 to 4i\sqrt{17}.

$$x=\frac{28+4\sqrt{17}i}{4}$$

Divide 28+4i\sqrt{17} by 4.

$$x=7+\sqrt{17}i$$

Now solve the equation x=\frac{28±4\sqrt{17}i}{4} when ± is minus. Subtract 4i\sqrt{17} from 28.

$$x=\frac{-4\sqrt{17}i+28}{4}$$

Divide 28-4i\sqrt{17} by 4.

$$x=-\sqrt{17}i+7$$

The equation is now solved.

$$x=7+\sqrt{17}i x=-\sqrt{17}i+7$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(14-x\right)^{2}.

$$x^{2}+196-28x+x^{2}=8^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+196-28x=8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$2x^{2}+196-28x=64$$

Subtract 196 from both sides.

$$2x^{2}-28x=64-196$$

Subtract 196 from 64 to get -132.

$$2x^{2}-28x=-132$$

Divide both sides by 2.

$$\frac{2x^{2}-28x}{2}=-\frac{132}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{28}{2}\right)x=-\frac{132}{2}$$

Divide -28 by 2.

$$x^{2}-14x=-\frac{132}{2}$$

Divide -132 by 2.

$$x^{2}-14x=-66$$

Divide -14, the coefficient of the x term, by 2 to get -7. Then add the square of -7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-14x+\left(-7\right)^{2}=-66+\left(-7\right)^{2}$$

Square -7.

$$x^{2}-14x+49=-66+49$$

Add -66 to 49.

$$x^{2}-14x+49=-17$$

Factor x^{2}-14x+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-7\right)^{2}=-17$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-7\right)^{2}}=\sqrt{-17}$$

Simplify.

$$x-7=\sqrt{17}i x-7=-\sqrt{17}i$$

Add 7 to both sides of the equation.

$$x=7+\sqrt{17}i x=-\sqrt{17}i+7$$